Problem: $f(x)=13-x$ $h(x)=x^2-6x-18$ Write $(f\circ h)(x)$ as an expression in terms of $x$. $(f\circ h)(x)=$
First, let's write $(f\circ h)(x)$ as $f(h(x))$. Next, we write $h(x)$ as the input to function $f$. $f({h(x)})=13-{h(x)}$ Since $h(x)=x^2-6x-18$, this becomes: $\begin{aligned} f({h(x)})&=13-({x^2-6x-18})\\ \\ &=13-x^2+6x+18\\ \\ &=31+6x-x^2\\ \\ \end{aligned}$ Note: We simplified the result to obtain a nicer expression, but this is not necessary. The answer: $(f\circ h)(x)=31+6x-x^2$